Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.
Answers
By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a
Here, b is any positive integer .
Let a be an arbitrary positive integer, then corresponding to the positive integers a and 6, there exist, non-negative integers q and r such that
a = 6q + r, where 0 ≤ r < 6
Squaring both the sides using (a + b)^2 = a^2 + 2ab + b^2
⇒ a^2 = (6q + r)^2 = 36q^2 + r^2 + 12qr
⇒ a^2 = 6(6q^2 + 2qr) + r^2
where,0 ≤ r < 6
Case I When r = 0 we get
⇒ a^2 = 6(6q^2) = 6m
where, m = 6q^2 is an integer.
Case II when r = 1 we get
⇒ a^2 = 6(6q^2 + 2q) + 1 = 6m + 1
where, m = (6q^2 + 2q) is an integer.
Case III When r = 2 we get
⇒ a^2 = 6(6q^2 + 4q) + 4 = 6m + 4
where, m = (6q^2 + 4q) is an integer.
Case IV When r = 3 we get
⇒ a^2 = 6(6q^2 + 6q) + 9
⇒ a^2 = 6(6q^2 + 6a) + 6 + 3
⇒ a^2 = 6(6q^2 + 6q + 1) + 3 = 6m + 3
where, m = (6q + 6q + 1) is an integer.
Case V when r = 4 we get
⇒ a^2 = 6(6q^2 + 8q) + 16
⇒ a^2 = 6(6q^2 + 10q) + 24 + 1
⇒ a^2 = 6(6q^2 + 8q + 2) + 4 = 6m + 4
where, m = (6q^2 + 8q + 2) is an integer
Case VI When r = 5 we get
⇒ a^2 = 6(6q^2 + 10q) + 25
⇒ a^2 = 6(6q^2 + 10q) + 24 + 1
⇒ a^2 = 6(6q^2 + 10q + 4) + 1 = 6m + 1
where, m = (6q^2 + 10q + 1) + 1 is an integer.
Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.
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